最长公共子串 LCS 模板题（Common Subsequence ）

￡神魔★判官ぃ 2023-06-01 06:12 12阅读 0赞

*  最长公共子串 LCS 模板题（Common Subsequence ）
    
     *  求最长公共子序列
     *  模板

# 最长公共子串 LCS 模板题（Common Subsequence ） #

[hdu-1159][]  
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.  
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.  
**Input**  
abcfbc abfcab  
programming contest  
abcd mnp  
Output  
4  
2  
0  
**Sample Input**  
abcfbc abfcab  
programming contest  
abcd mnp  
**Sample Output**  
4  
2  
0

## 求最长公共子序列 ##

## 模板 ##

#include <algorithm> #include <iostream> #include <cstring> #define N 1003 using namespace std; char a[N], b[N]; int dp[N][N]; int main() { while (scanf("%s%s", a, b) != EOF) { memset(dp, 0, sizeof(dp)); int lena = strlen(a); int lenb = strlen(b); for (int i = 1; i <= lena; i++) { for (int j = 1; j <= lenb; j++) { if(a[i-1]==b[j-1]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]); } } cout<<dp[lena][lenb]<<endl; } }

转载于:https://www.cnblogs.com/tttfu/p/11344958.html

[hdu-1159]: http://acm.hdu.edu.cn/showproblem.php?pid=1159